An equation is used to show two expressions that are equal to each other.

Solving linear equations where brackets are used

Maths

In this unit, pupils learn how to rearrange an equation to find a specific value that satisfies the equation. In expressions and formulae, pupils use their understanding of rearrangement to change the subject of a formula.

In the prior unit, pupils saw that an arithmetic sequence can be represented graphically and that this is a linear relationship. Any point on the line satisfies this relationship. In this unit, pupils learn how to find a value that satisfies the relationship when the second value is known, without the aid of the graph.

Solving linear equations

$$8(x+5)=5(2x-4)$$ is an example of {{}}.

$$9y-3=45$$, therefore $$9y=48$$ and $$y={48\over9}$$

$$9y-27=45$$, therefore $$9y=72$$ and $$y=8$$

$$9y-27=45$$, therefore $$9y=18$$ and $$y=2$$

Which of these show a correct method and solution for the equation $$9(y-3)=45$$?

The solution to $$8(x+5)=5(2x-4)$$ is $$x=$${{}}.

Which is the most efficient start to solving the equation $${5\over4}(3x+1)=20$$?

The solution to the equation $${5\over4}(3x+1)=20$$ is $$x=$${{}}.

The solution to $$8y+20=4$$ is $$y=$${{}}.

This lesson will soon be taken down.

Solving linear equations where brackets are used

Lesson outcome